#11**+5 **

**Hi Sabi92, **

**Welcome to the Web2 forum **

This is a very strange place to put your question.

Just put it on a new post next time. :)

I really like that fact that you have had a go at it yourself and that you have shown us your work :)

$$\\a(2b^2-c)+b(c-2b^2)$$

what you need to notice here is that the brackets contain the same things except the signs are different.

I am going to look at the second bracket. I could just as easily chosen the first bracket :)

$$c-2b^2 = -1(-c+2b^2)=-1(2b^2-c)$$

so

$$+b(c-2b^2)=+b*-1(2b^2-c) = -b(2b^2-c)$$

so

$$\\a(2b^2-c)+b(c-2b^2)\\\\

=a(2b^2-c)-b(-c+2b^2)\\\\

=a\textcolor[rgb]{0,1,0}{(2b^2-c)}-b\textcolor[rgb]{0,1,0}{(2b^2-c)}\\\\

$Now you have a lots of the green stuff minus b lots of the green stuff$\\\\

$which equals (a-b) lots of the green stuff$\\\\

=(a-b)(2b^2-c)\\\\$$

Melody
Jun 7, 2015

#3**+5 **

** i dont know how to solve can someone help me**

6x=x^2

x=?

$$\small{\text{

$

\begin{array}{rcl}

6x & = & x^2 \\

x^2 - 6x &=& 0 \\

\underbrace{x}_{=0}(\underbrace{x-6}_{=0}) &=& 0 \\

\end{array}

$

}}$$

1. x = 0

2. x-6 = 0 | +6

x = 6

heureka
Nov 17, 2014

#6**0 **

oops im sorry i wrote a wrong question the real question is

6x=2x^2

x=?

so how can i solve it?

Guest Nov 17, 2014

#7**+5 **

** i dont know how to solve can someone help me**

6x=2x^2

x=?

$$\small{\text{

$

\begin{array}{rcl}

6x & = & 2x^2 \\

2x^2 - 6x &=& 0 \\

\underbrace{x}_{=0}(\underbrace{2x-6}_{=0}) &=& 0 \\

\end{array}

$

}}$$

1. x = 0

2. 2x-6 = 0 | +6

2x = 6 | : 2

x = 6 /2

x = 3

heureka
Nov 17, 2014

#8**+5 **

It is similar to what Heureka did.

$$\\6x=2x^2\\

3x^2-6x=0\\

3x(x-2)=0$$

you should be able to finish it :)

Melody
Nov 17, 2014

#10**0 **

i need to do a factorization but i dont get the right answer.help please.

a(2b^2-c)+b(c-2b^2)

that's what i did:

2ab^2-ac+bc-2b^3=a(2b^2-c)+b(c-2b^2)

now what?

sabi92
Jun 7, 2015

#11**+5 **

Best Answer

**Hi Sabi92, **

**Welcome to the Web2 forum **

This is a very strange place to put your question.

Just put it on a new post next time. :)

I really like that fact that you have had a go at it yourself and that you have shown us your work :)

$$\\a(2b^2-c)+b(c-2b^2)$$

what you need to notice here is that the brackets contain the same things except the signs are different.

I am going to look at the second bracket. I could just as easily chosen the first bracket :)

$$c-2b^2 = -1(-c+2b^2)=-1(2b^2-c)$$

so

$$+b(c-2b^2)=+b*-1(2b^2-c) = -b(2b^2-c)$$

so

$$\\a(2b^2-c)+b(c-2b^2)\\\\

=a(2b^2-c)-b(-c+2b^2)\\\\

=a\textcolor[rgb]{0,1,0}{(2b^2-c)}-b\textcolor[rgb]{0,1,0}{(2b^2-c)}\\\\

$Now you have a lots of the green stuff minus b lots of the green stuff$\\\\

$which equals (a-b) lots of the green stuff$\\\\

=(a-b)(2b^2-c)\\\\$$

Melody
Jun 7, 2015